tensorflow的simple linear regression

例子出自Hackers guide to machine learning with python

使用的数据文件非常小，只有两列，分别为Speed和Stopping Distances of Cars. 数据文件 其中的cars

代码如下：

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1. # coding: utf-8
   
2. 
   
3. import pandas as pd
   
4. import tensorflow as tf
   
5. import matplotlib
   
6. matplotlib.use('Qt5Agg')
   
7. import matplotlib.pyplot as plt
   
8. import seaborn as sns
   
9. 
   
10. sns.set(style='whitegrid', palette='muted', font_scale=1.5)
    
11. 
    
12. data_csv=pd.read_csv('datasets/cars.csv', usecols=['speed', 'dist'])
    
13. print(data_csv.head())
    
14. 
    
15. target = data_csv['dist']
    
16. speed = data_csv['speed']
    
17. 
    
18. #plot datasets
    
19. sns.scatterplot(speed, target);
    
20. plt.xlabel("speed")
    
21. plt.ylabel("stopping distance");
    
22. 
    
23. 
    
24. dataset=tf.data.Dataset.from_tensor_slices((speed.values, target.values))
    
25. for feat, targ in dataset.take(5):
    
26.     print('Feature: {}, Target: {}'.format(feat, targ))
    
27.     
    
28. 
    
29. from tensorflow.keras import layers
    
30. lin_reg=tf.keras.Sequential([layers.Dense(1, activation='linear',input_shape=[1]),])
    
31. optimizer = tf.keras.optimizers.RMSprop(0.001)
    
32. 
    
33. lin_reg.compile(loss='mse', optimizer=optimizer, metrics=['mse'])
    
34. history = lin_reg.fit(x=speed, y=target, shuffle=True, epochs=1000, validation_split=0.2, verbose=0)
    
35. 
    
36. def plot_error(history):
    
37.     hist = pd.DataFrame(history.history)
    
38.     hist['epoch'] = history.epoch
    
39.     
    
40.     plt.figure()
    
41.     plt.xlabel('Epoch')
    
42.     plt.ylabel('Mean Square Error')
    
43.     plt.plot(hist['epoch'], hist['mse'],
    
44.                 label='Train Error')
    
45.     plt.plot(hist['epoch'], hist['val_mse'],
    
46.                 label = 'Val Error')
    
47.     
    
48.     plt.legend()
    
49.     plt.show()
    
50. 
    
51. plot_error(history)
    
52. 
    
53. print(lin_reg.summary())
    
54. #get weights
    
55. weights = lin_reg.get_layer("dense").get_weights()
    
56. intercept = weights[0][0][0]
    
57. slope = weights[1][0]
    
58. 
    
59. print('weigths: {}'.format(weights))
    
60. print('intercept: {}'.format(intercept))
    
61. print('slope: {}'.format(slope))
    
62. 
    
63. 
    
64. #build simple neural network
    
65. def build_neural_net():
    
66.     net = tf.keras.Sequential([
    
67.         layers.Dense(32, activation='relu', input_shape=[1]),
    
68.         layers.Dense(16, activation='relu'),
    
69.         layers.Dense(1),
    
70.     ])
    
71.     
    
72.     optimizer = tf.keras.optimizers.RMSprop(0.001)
    
73.     
    
74.     net.compile(loss='mse',
    
75.                     optimizer=optimizer,
    
76.                     metrics=['mse', 'accuracy'])
    
77.     
    
78.     return net
    
79. 
    
80. net = build_neural_net()
    
81. 
    
82. history = net.fit(
    
83.     x=speed,
    
84.     y=target,
    
85.     shuffle=True,
    
86.     epochs=1000,
    
87.     validation_split=0.2,
    
88.     verbose=0
    
89. )
    
90. 
    
91. plot_error(history)
    
92. 
    
93. #stop training early
    
94. early_stop = tf.keras.callbacks.EarlyStopping(
    
95.     monitor='val_loss',
    
96.     patience=10
    
97. )
    
98. 
    
99. 
    
100. net = build_neural_net()
     
101. 
     
102. #simple neural network summary
     
103. print(net.summary())
     
104. print(net.weights)
     
105. 
     
106. 
     
107. history = net.fit(
     
108.     x=speed,
     
109.     y=target,
     
110.     shuffle=True,
     
111.     epochs=1000,
     
112.     validation_split=0.2,
     
113.     verbose=0,
     
114.     callbacks=[early_stop]
     
115.     )
     
116. 
     
117. plot_error(history)
     
118. 
     
119. #save & restore model
     
120. net.save('simple_net.h5')
     
121. simple_net = tf.keras.models.load_model('simple_net.h5')
     
122. print(simple_net.summary())

simple linear regression得出的slope 和intercept
试过几次，差距很大， 为什么呢？

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1. intercept: 2.5482685565948486
   
2. slope: 0.8580179810523987
   
3. 
   
4. intercept: 0.6354206800460815
   
5. slope: 1.985124111175537
   
6. 那么得出的equation就会分别是
   
7. y = 0.8580179810523987 * x + 2.5482685565948486
   
8. y = 1.985124111175537 * x + 0.6354206800460815